Problem VI asks to "find a number which is equal to its square root plus twice its cube root." Denoting the positive number by x6, we have to solve the equation x6= x3 + 2x2. Dividing by x2 we get the equivalent equation x4= x + 2, a quartic indeed. Cardano subtracts 1 from each side and reaches x4 - 1= x + 1, thus (x4 - 1)/(x + 1)= 1. Consequently x3 - x2+ x - 1= 1, that is to say x3 + x = x2 + 2. He ascertains that $$x=\root 3\of{\sqrt{83\over 108}+{47\over 54}}-\root 3\of{\sqrt{83\over 108}-{47\over 54}}+{1\over 3}$$is a solution, without providing the details. Let us check whether this is a solution, using for that purpose the Cardano-Tartaglia method.

A natural question, addressed in Ars Magna, is what happens if we follow Ferrari's procedure: We start with (x2)2 = x + 2 and introduce a new variable b. We observe that (x2 + b)2 = 2bx2 + x + b2 + 2. In order to have a perfect square we need to impose the condition 1/4 = 2b(b2 + 2), equivalent to demanding that the discriminant of 2bx2 + x + b2 + 2 is zero. Thus b3 + 2b= 1/8. Cardano ascertains that $$b=\root 3\of{{1\over 16}+\sqrt{2,075\over 6,912}}-\root 3\of{-{1\over 16}+\sqrt{2,075\over 6,912}}$$is a solution. We would have to replace this value of b, call it b1, in (x2 + b)2= 2bx2 + x + b2 + 2 knowing beforehand that the expression to the right is also a perfect square. In other words, it is necessary to solve (x2 + b1)2 = 2b1(x+ 1/4b1)2 or, equivalently, the pair of quadratic equations $$x^2+b_1=\sqrt{2b_1}\left(x+{1\over 4b_1}\right), \qquad x^2+b_1=-\sqrt{2b_1}\left(x+{1\over 4b_1}\right).$$The solutions of the first quadratic are $$x=\sqrt{b_1\over 2} \pm\sqrt{{1\over 2\sqrt{2b_1}}-{b_1\over 2}}.$$

Using the decimal approximation 0.062379 to b1 we get x = 1.35321 and x = -0.999998. The first is a decimal approximation to the solution found before, while the other root is a decimal approximation to -1. But we are not interested in the latter because we are seeking only positive solutions to the problem (from the very beginning we ruled out the trivial solutions of the equation x6 = x3 + 2x2, namely 0 and -1). It should be noted that the quadratic equation \( x^2 +b_1 = -\sqrt{2b_1} \left( x+\frac{1}{4b_1} \right) \) has complex roots.

Harald Helfgott (University of Bristol) and Michel Helfgott (East Tennessee State University), "A Modern Vision of the Work of Cardano and Ferrari on Quartics - Different Strategies," Loci (February 2010), DOI:10.4169/loci003312